$\lim_{x\to\infty}\dfrac{-3x^3+2x}{4x+8}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{9}{4}$ (Choice B) B $-\dfrac{3}{4}$ (Choice C) C $0$ (Choice D) D $-\infty$
Answer: $\lim_{x\to\infty} -3x^3+2x=-\infty$ and $\lim_{x\to\infty} 4x+8=\infty$, so $\lim_{x\to\infty}\dfrac{-3x^3+2x}{4x+8}$ results in the indeterminate form $\dfrac{-\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{-3x^3+2x}{4x+8} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[-3x^3+2x\right]}{\dfrac{d}{dx}[4x+8]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{-9x^2+2}{4} \\\\ &=-\infty \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[-3x^3+2x\right]}{\dfrac{d}{dx}[4x+8]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{-3x^3+2x}{4x+8}=-\infty$.